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3 years ago

What is the gravitational force between mars and Phobos

Physics

1 answer:

alina1380 [7]3 years ago

8 0

Answer:

$F=5.16\times 10^{15}\ N$

Explanation:

We have,

Mass of Mars is, $m_M=6.42\times 10^{23}\ kg$

Mass of its moon Phobos, $m_P=1.06\times 10^{16}\ kg$

Distance between Mars and Phobos, d = 9378 km

It is required to find the gravitational force between Mars and Phobos. The force between two masses is given by

$F=G\dfrac{m_Mm_P}{d^2}$

Plugging all values, we get :

$F=6.67\times 10^{-11}\times \dfrac{6.42\times 10^{23}\times 1.06\times 10^{16}}{(9378\times 10^3)^2}\\\\F=5.16\times 10^{15}\ N$

So, the gravitational force is $5.16\times 10^{15}\ N$ .

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Because many fuels are fossil fuels they take millions of years to form and the known reserves are being used much faster than the new ones being made

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Compare the forces in a small nucleus to the forces in a large nucleus

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4 years ago

0.5-lbm of a saturated vapor is converted to asaturated liquid by being cooled in a weighted piston-cylinder device maintained a

klio [65]

Answer:

The boiling point temperature of this substance when its pressure is 60 psia is 480.275 R

Explanation:

Given the data in the question;

Using the Clapeyron equation

$(\frac{dP}{dT} )_{sat } = \frac{h_{fg}}{Tv_{fg}}$

$(\frac{dP}{dT} )_{sat } = \frac{\frac{H_{fg}}{m} }{T\frac{V_{fg}}{m} }$

where $h_{fg$ is the change in enthalpy of saturated vapor to saturated liquid ( 250 Btu

T is the temperature ( 15 + 460 )R

m is the mass of water ( 0.5 Ibm )

$V_{fg$ is specific volume ( 1.5 ft³ )

we substitute

$(\frac{dP}{dT} )_{sat } =( \frac{250Btu\frac{778Ibf-ft}{Btu} }{0.5})$ / $( (15+460)\frac{1.5}{0.5})$

$(\frac{dP}{dT} )_{sat } =$ 272.98 Ibf-ft²/R

Now,

$(\frac{dP}{dT} )_{sat } =$ $(\frac{P_2 - P_1}{T_2 - T_1})_{sat$

where P₁ is the initial pressure ( 50 psia )

P₂ is the final pressure ( 60 psia )

T₁ is the initial temperature ( 15 + 460 )R

T₂ is the final temperature = ?

we substitute;

$T_2$ $= ( 15 + 460 ) + \frac{(60-50)psia(\frac{144in^2}{ft^2}) }{272.98}$

$T_2 = 475 + 5.2751\\$

$T_2 =$ 480.275 R

Therefore, boiling point temperature of this substance when its pressure is 60 psia is 480.275 R

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3 years ago

A 1-kg iron frying pan is placed on a stove. The pan increases from 20°C to 250°

timurjin [86]

Answer;

The temperature change for the second pan will be lower compared to the temperature change of the first pan

Explanation;

-The quantity of heat is given by multiplying mass by specific heat and by temperature change.

That is; Q = mcΔT

This means; the quantity of heat depends on the mass, specific heat capacity of a substance and also the change in temperature.

-Maintaining the same quantity of heat, with another pan of the same mass and greater specific heat capacity would mean that the change in temperature would be much less lower.

8 0

4 years ago

Read 2 more answers

An object has a starting velocity of 20 m/s. If it accelerates at a rate of 2 m/s' for 3 seconds,

Ghella [55]

Answer:

26 m/s

69 m

Explanation:

Given:

v₀ = 20 m/s

a = 2 m/s²

t = 3 s

Find: v and Δx

v = at + v₀

v = (2 m/s²) (3 s) + 20 m/s

v = 26 m/s

Δx = v₀ t + ½ at²

Δx = (20 m/s) (3 s) + ½ (2 m/s²) (3 s)²

Δx = 69 m

3 0

3 years ago